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For me the fourier transform of a measure $\mu\in\mathcal{S}'(\mathbb{R})$ is defined by $\hat{\mu}(\varphi)=\mu(\hat{\varphi})$ where $\varphi\in\mathcal{S}(\mathbb{R})$.

My question is: if one has $\int_\mathbb{R}\operatorname{e}^{-\lambda s}d\mu(s)=f(\lambda)$ for all $\Re(\lambda)>0$, and if $f$ has a continuous prolonging on $\Re(z)=0$ with $t\mapsto f(t)\in \operatorname{L}^1_{loc}(\mathbb{R})$, can we conclude that the Fourier transform of $\mu$ is $t\mapsto f(2i\pi t)$?

My question comes from arithmetic context: with $\mu=\sum_{p\in\mathbb{P}}\frac{1}{p}\delta_{\log{p}}$ we have easely that $\int_\mathbb{R}e^{-\lambda t}d\mu(t)=\zeta_\mathbb{P}(1+\lambda)$ and I'd like to conclude that $\hat{\mu}=\zeta_{\mathbb{P}}(1+2i\pi t)$. I know a specific proof of that but I'd like to know if it were a general thing.

Gabriel Soranzo
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    If $\mu$ isn't finite, $\int_\mathbb{R} e^{-2\pi its},d\mu(t)$ doesn't exist in the Lebesgue sense. If that integral can be sensibly defined, then $\hat{\mu}$ should be $L_f$, but I don't know which conditions guarantee that. – Daniel Fischer Jan 05 '14 at 13:57
  • Thanks, I changed a little my question – Gabriel Soranzo Jan 05 '14 at 14:37
  • Your measure has support in $(0,\infty)$, hence $\mu_\xi \colon f \mapsto \int f(s)e^{-\xi s},d\mu(s)$ converges to $\mu$ in the sense of tempered distributions. Thus $\widehat{\mu_\xi} \xrightarrow{\xi\to 0} \widehat{\mu}$ in the same sense. so the question is whether $t\mapsto \zeta_{\mathbb{P}}(1+\xi + 2\pi it)$ converges to $t\mapsto \zeta_{\mathbb{P}}(1+ 2\pi i t)$ in the sense of tempered distributions. I'm pretty sure it does, but one would need to look at $\zeta_\mathbb{P}$ to prove it. – Daniel Fischer Jan 05 '14 at 15:39
  • Yes, it is the proof I have: $\zeta_\mathbb{P}(1+\xi+2i\pi t)$ converges to $\zeta_\mathbb{P}(1+2i\pi t)$ in $\mathcal{D}'(\mathbb{R})$ (it's enough). I found so clear the identity $\int_\mathbb{R}e^{-\lambda t}d\mu(t)=\zeta_\mathbb{P}(1+\lambda)$ that I hoped a short argument can conclude without this technic of approximation: this was the motivation of my question – Gabriel Soranzo Jan 05 '14 at 20:22
  • Unfortunately, I think one can't. One thing that makes it hairy is that $\zeta_\mathbb{P}$ has a pole in $1$ (I think, or is it even not an isolated singularity of $\zeta_\mathbb{P}$?), so $t\mapsto \zeta_\mathbb{P}(1+2\pi it)$ is not locally integrable, and you must consider the principal value integral in the limit. I think (but don't know) the behaviour for large $\lvert t\rvert$ is harmless enough that you can argue with locally uniform (and dominated) convergence there, so $\lvert t\rvert$ small would be the problematic part. – Daniel Fischer Jan 05 '14 at 20:32
  • As far as I know, element of $\mathscr S'$ is called tempered distribution not a measure which related to measurable space. – Zbigniew Dec 07 '15 at 15:58

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